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Math Help - More questions on parametic curves in 2D

  1. #1
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    More questions on parametic curves in 2D

    Hi

    My question has two parts:
    Using the following two equation, x=lnU & y=ln(\frac{2}{U} find:
    1) calculate \frac{d^2y}{dx^2}

    This is what i have done:

    find \frac{dy}{dx} first

    \frac{dy}{dx} = \frac{dy}{dU} \frac{1}{\frac{dx}{dU}}
    \frac{dy}{dx} = \frac{-2U^{-2}}{2U^{-1}} \frac{1}{\frac{1}{U}}
    \frac{dy}{dx} = \frac{-u}{u}
    \frac{dy}{dx} = -1

    \frac{d^2y}{dx^2} = -U

    where have a gone wrong?

    2) eliminate the parameter to give a direct relation between x and y:

    tried this way but was wrong.
    e^x = U

    y=ln(\frac{2}{e^x}

    answer in book is y=ln2-x
    P.S
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  2. #2
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    \displaystyle \log\left( \frac{a}{b}\right)=\log a-\log b,

    any base.
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  3. #3
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    Hello, Paymemoney!

    Given: . \begin{Bmatrix} x &=& \ln u \\ y &=& \ln(\frac{2}{u}) \end{bmatrix}

    1) Find: . \dfrac{d^2y}{dx^2}

    Note that: . u > 0

    We have: . y \:=\:\ln2 - \ln u \quad\Rightarrow\quad \dfrac{dy}{du} \:=\:-\dfrac{1}{u}

    And: . x \:=\:\ln u \quad\Rightarrow\quad \dfrac{dx}{du} \:=\:\dfrac{1}{u}


    Hence: .  \displaystyle \frac{dy}{dx} \;=\;\frac{\frac{dy}{du}}{\frac{dx}{du}} \;=\;\frac{\text{-}\frac{1}{u}}{\frac{1}{u}} \;=\;-1

    Therefore: . \dfrac{d^2y}{dx^2} \;=\;0




    2) Eliminate the parameter to give a direct relation between \,x and \,y.

    Answer in book is: y\:=\:\ln2-x

    We have:. . \begin{array}{ccccccccc}<br />
x &=& \ln u & [1] \\<br />
y &=& \ln(\frac{2}{u}) & \Rightarrow & y &=& \ln 2 - \ln u & [2] \end{array}


    Substitute [1] into [2]: . y \;=\;\ln 2 - x
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